Let $g(x)=x^{^{\scriptsize\dfrac{4}{3}}}$. $g'(27)=$
Explanation: Let's first find the expression for $g'(x)$ and then evaluate it at $x=27$. The derivative of $g$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a fraction.) $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left(x^{^{\frac{4}{3}}}\right) \\\\ &=\dfrac{4}{3}x^{^{\frac{4}{3}-1}} \gray{\text{The power rule}} \\\\ &=\dfrac{4}{3}x^{^{\frac{1}{3}}} \end{aligned}$ So we found that $g'(x)=\dfrac{4}{3}x^{^{\frac{1}{3}}}$, which can also be written as $\dfrac{4}{3}\sqrt[3]{x}$ Now let's plug ${x=27}$ : $\begin{aligned} \dfrac43\sqrt[3]{{27}}&=\dfrac43\cdot 3 \\\\ &=4 \end{aligned}$ In conclusion, $g'(27)=4$.